You could use the mean (average), mode (most common), median (central value of the ordered set) as a single figure to represent the data. The mean is found by adding the calorific values and dividing by the number of data=9. I think that, without analysing the data more closely, in this case the mean would be best, because there is no mode, since each value is unique. If we work out the mean and median we may find they have similar values.

The data is already ordered and the median is the 5th value=607. The value 932 is an “outsider” (outlier) which will make the mean higher than 607. Nevertheless we can calculate the mean and compare it with the median.

The mean is 632.9 which is a lot higher than 607, and is actually higher than the penultimate value 632, and if presented as a general single value, it may mislead customers into thinking that the sandwiches have a high calorific value. We don’t know how many sandwiches of different types there are, and there may only be a few with high values. **So I would use the median 607 as a typical value for the number of calories.**

- All categories
- Pre-Algebra Answers 12.4k
- Algebra 1 Answers 25.4k
- Algebra 2 Answers 10.5k
- Geometry Answers 5.2k
- Trigonometry Answers 2.7k
- Calculus Answers 6.1k
- Statistics Answers 3k
- Word Problem Answers 10.2k
- Other Math Topics 6.8k

82,167 questions

86,657 answers

2,246 comments

76,258 users